I've been doing the puzzles on this site for about a month, and this one has me stumped. Please help. This is where I am stopped:

67x xxx x89
938 762 xx1
xx1 98x 7xx

1xx xxx 9xx
8x3 xxx 1x2
7x6 xxx xx8

x89 531 6xx
567 249 813
31x x7x x9x

Hello jimeg39!

When I get stuck it is usually because of a pair. Look for a 4/5 pair in r2c7 and r2c8. Then you can erase 4/5 from r3c9 and put a 6 there. You also have a hidden pair 2/9 in r6c2 and r6c5. Erase everything but the 2/9 from these spaces and you will see a 1 as the sole candidated in r1c5.

When I get stuck I look in the rows and columns for places where a number only appears twice and pencil in a box around it. That helps me to find the hidden pairs.

> When I get stuck it is usually because of a pair. I look in the rows and
> columns for places where a number only appears twice and pencil in a
> box around it. That helps me to find the hidden pairs.

Last month I posted on the subject of "Mandatory Pairs". This method will
usually solve medium and difficult puzzles without resort to candidate
profiles. With "hard" and "very hard" it can often take one a long way into
the puzzle and, without noticing it (!), will often resolve the hidden pairs
BEFORE they become a problem.

When Mandatory pairs fails to yield further, it is easy to convert it into
candidate profiles method - especially as the candidates are much fewer
in number. Of especial use is the feature that "mutual reception" in the
Mandatory Pairs method translates directly to candidate profiles and
clears hidden pairs without any seeking for them.

I used Mandatory Pairs for most of the Nov 5th solution but had to switch
to candidate profiles eventually. There the major advances were the
splitting of column 6, row 4, row 5 and row 6 into subsets of profiles
(eg hidden triplets) and "counting" whenever a row column or region
gets up to six cells completed. With a row/colum looking for two of the
missing digits in the column/row at right angles to each blank cell can
often provide the one vital clue to resolve the puzzle - helpfully before
compiling the full set of candidate profiles in a lot of cases!

It is not always necessary but one useful tip is to record at the edge of
the puzzle the profile for each row/column - split into subsets where this
is applicable. For instance row 6 split into (345) and (29). Marking these
at the edge of the puzzle saved having to work out the situation each
time when looking at the cells involved from a different perspective.

I do not know for sure but I suspect that the 3x3 nature of the standard
puzzle means that pairs are more common in reaching a solution. With
the "either ON or OFF" nature of binary pairs, any sytem of solution
which narrows down the result to A or B (with no possibility of anything
else!) has the magnificient feature of "If it is not A then it MUST be B"

If three cells have pair links 234, 24, 3 one can immediately exclude the
3 from the first pair - leaving a mutual reception - AND set 3 in the third
cell. However, one must be careful as the rules for candidate profiles
include "exhaustive checks" (a single candidate MUST be the one) whereas
Mandatory Pairs do not have such a feature. Their power is that if one
can prove that a cell does not have a value then its pair must be true
where candidate profiles would need to reduce the number of occurrences
in the row/column/region to just one in order to reach the same result.

Congratulations on recognising the power of pairs.
Have fun discovering ways to harness that knowledge!

I also struggled with this Nov 5 beast and finally got it using trial-and error (a version of it) by looking for pairs and trying one until an impossible situation arose, or the puzzle was solved. I got as far as the following using logic before resorting to this method.
670000089
938762001
001980706
100008900
803007102
706000008
089531600
567249813
310876090

the "hint" in the draw function suggested a 7 at r4c9, but I cannot see why this "has" to be there (in other words, I know it's correct but what's the compelling logic?) Help!

You should look at Row 4. There is a triplet of 2,4,5 in the cells r4c2, r4c3 and r4c5. This means you can eliminate all other Nrs 2,4 and 5 from other cells in Row 4. Since the only possible numbers for cell r4c9 are 4,5 and 7, this leaves Nr 7 as the only possibility for that cell.

[quote="jimeg39"]I've been doing the puzzles on this site for about a month, and this one has me stumped. Please help. This is where I am stopped:

67x xxx x89
938 762 xx1
xx1 98x 7xx

1xx xxx 9xx
8x3 xxx 1x2
7x6 xxx xx8

x89 531 6xx
567 249 813
31x x7x x9x[/quote]

In r2 numbers 4 and 5 are missing; r2c7 can be 4 (in which case r2c8 turns 5), or r2c7 can be 5 (in which case r2c8 turns 4); in either case r3c9 can not be 4 or 5; so r3c9 can only be 6. Hope this helps; I have not finished the puzzle; but I had exactly the same pattern as you (plus the 6 in r3c9); then I got stuck.
PS: I can not find the three triplets Chris mentions (because r8c5 = 4, so r4c5 can not be 2, 4 or 5, only 2 or 5).

I was the person who mentioned the triplet of 2,4,5 in Row 4. Let me explain about triplets ( or even quadruplets - 4 numbers ). The fact there is a triplet in a Row does NOT mean that the cells involved have to contain all 3 numbers - it just means that there are a total of 3 separate numbers that are solely self-contained within 3 separate cells. For example, you may have a triplet of 5,6,7 where one cell contains the possibilities of 5,6 - the next cell contains 6,7 - and the third cell contains 5,7. This is still defined as a triplet because no matter how you look at the Row, these three numbers must somehow appear in these three cells. The same logic applies for quadruplets where there are 4 numbers involved in 4 cells. For example you could have a quadruplet of 1,4 - 4,5 - 4,5,6 and 1,56.

In any event you are right when you say r4c5 can only contain 2 or 5. But the fact remains a triplet appears in r4c2, r4c3 and r4c5 because the numbers 2,4 and 5 must appear only in those three cells.

I received an e-mail asking about this puzzle and so I had an
opportunity to rework it - and to document it at the same time.

Using mandatory Pairs, I got to exactly the same position as
the forum participant who first raised this topic. From that point
I needed to resort to using Candidate Profiles. The detail follows.

+++

November 5th Solution

The first steps use Mandatory Pairs.
Thus not all moves set values.
Notes:
MP means mark Mandatory Pair (ie that the
value stated can exist in only two cells
of the nine in the box. These notes do
not state which two - find by inspection!
M/R means Mutual Reception (ie that two of
the cells have the same eligible Mandatory
Pairs in them - thus excluding all else.

A) My first procedure is to check the "wide columns"
using each confirmed cell in turn to determine if
it has any immediate implications.

B) The next stage is to check any lines with six or
more resolved cells to determine if any of the
unresolved cells in them have "sole" qualities.

08 - set 7 in r8c3 (sole candidate of 3,7 in row 8)
09 - set 3 in r8c9 (last cell in row)

C) The third stage is to check "broad rows" as in (A)

09a - MP 1 in box 2
10 - set 9 in r1c9 (r3c9 eliminated as 9 in r3c4)
(NB This could have been done after 7a. It is
not uncommon to leave something like this and
to find it again later. The important thing is
that it be cleared before moving to Candidate
Profiles!)
10a - MP 6 in box 3 (possible now that r1c9 is set)
11 - set 6 in r1c1 (6 pair in box 3 eliminates r3c1)
11a - MP 8 in box 5
11b - MP 5 in box 9
11c - MP 7 in box 9
11d - MP 9 in box 7

D) A similar check as in (B) is undertaken here but no
new information emerges.

E) The next stage is an overall check for each digit in
turn. I normally work in sequence 1-9 but revisit a
digit if setting a higher digit opens up potential
for a lower digit to be set.

11e - Note that seven of digit 1 are set and the other
two are constrained to pairs. No action possible.
11f - No progress with 2 or 3.
11g - Note that grid contains only one 4 - no progress!
12 - set 7 in r6c1 (eliminate r4c3 as 7 in r8c3)
12a - MP 7 in box 5
13 - set 7 in r3c7 (reasoning below)
NB: The 7 in box 6 must be in col 8 or col 9. Box
9 has a MP for 7 in r7c8 and r7c9. Thus the 7 in
col 7 must be in box 3 - or the same thing can be
expressed as the 7 in box 3 must be in col 7.
The general rule is that where two boxes in a
broad column or a broad row have a digit that is
constrained to the same two columns (or rows) then
that digit must be in the third column/row in the
third box of the broad column/row. Very often this
is observable by seeing the MPs but sometimes the
constraint (as here) is more subtle. That is why
this stage is so important - even if often it does
not lead to anything new. One can pick up on any
earlier omissions - and maybe spot a subtlety.
13a - MP 2 in box 3 (a revisit after the 7 set in box 3)
13b - MP 3 in box 3 (similar to above!)
13c - Notice M/R in r1c7 and r3c8 and eliminate 6
14 - set 6 in r3c9 (consequent on above)
14a - Notice M/R of 4/5 sole possibility in box 3
14b - MP 4 in box 3
14c - MP 5 in box 3
15 - Set 9 in r2c1 (sole position once above M/R noted)
16 - set 9 in r7c3 (9 in r7c1 eliminated)
16a - MP 9 in box 4
16b - Notice M/R 2,4 in box 7
16c - MP 2 in box 7
16d - MP 4 in box 7
16e - MP 6 in box 6

F) This is the point at which one needs to consider the
transition to Candidate Profiles - although there is
one possible intermediate stage. This involves setting
out the "missing" digits for each row/column at the end
of it. A simple 'cross-wise' check of the intersecting
column/row may reveal a sole candidate.

On this occasion there is nothing to be gained directly
from the "Missing" profiles - largely because of the
inability to set any digit 4 within the puzzle.
However the "Missing" profiles are very useful for
deriving the Candidate Profiles - so not all was in vain.

H) Setting the Candidate profiles

This process is simplified by tackling first the rows and
columns with the fewest members of their "Missing Profile"
and checking which can be eliminated for each cell in the
row/column. It should be noted that any "Mutual Reception"
cases should be retained with the M/R digits and not be
allowed to gain extra candidates because of the profile
of the Row or Column.

When setting the profiles for each row/column, it is
important to look out for any opportunities to group
the overall profiles. This has been done already for
row 9 where the (68) and (245) form two sub-groups but
it is possible for these to come to light when setting
the profiles for individual cells within a row/column.

Having derived the profiles, it is now important to
scan them for congruence and for compatibility with
the already derived Mandatory Pairs.

a) Congruence is a check that any set of candidate
profiles within a row, column or box contains
only as many distinct values as the number of
unresolved cells in the said row/column/box.
Congruence must be checked for sub-groups as
well as for the whole row/column/box. If a
sub-group has been identified, it is possible
that elimination may be required from the OTHER
subgroup in order to restore congruence.

b) If a box contains a Mandatory Pair and that Pair
occupy cells in a straight line (row/column) they
must be the ONLY cells in that row/column that
contain the said digit and it may be removed from
any other cell in the same row/column.

NB: Any elimination of a candidate should be followed
by a re-check on congruence and possible emergence
of a sub-group not previously identified.

This provides the solution.
It took much longer to document it than to do it.
Looking back at my original workings, I find that
I have managed to solve one more cell this time
before reverting to candidate profiles.

Given the subtle interactions, there is no way
that this one could have been solved by Mandatory
Pairs alone - but they certainly helped!

I don't do full candidate profiles and therefore don't see such triplets etc.
Mostly this isn't a problem because such a triplet is usually accompanied by a pair which I see (first) instead.
But in this case I am stymied at 16f: no pairs apparent.

However there is a Glassman pan to get round this; it is based at the shared triplet r4b4.
The rest of box4 must have the same values as the rest of row4.

Initially it would appear inconclusive because the 3/6/7/8 of the pan's bowl and the 9 of the pan's handle leave the sixth value unconstrained.
But it does work, in two stages:
You can't fit 3/6/7/8 into the handle at col5 (so r4c5 apparently holds the sixth value).
Thus only four cells are available in the handle to accomodate 3/6/7/8. This is conclusive in the case of col9 which already has 3/6/8.

I think it pretty well much falls out then, so maybe you didn't need to go to full candidate profiles after all.

> I think it pretty well much falls out then, so maybe you
> didn't need to go to full candidate profiles after all.

Probably not.
That is one of the challenges when using a method like Mandatory
Pairs and one gets "stuck". There is the option to move to full
Candidate Profiles but, in a way, that is admitting defeat and on
occasion is proven to be a premature or unnecessary step.

My spatial vision is not overly good and I admit that I have not
really taken on board the "pan" syndromes. I have expressed
previously my amazement at the ability of our Denver friend to
"see" patterns in the initial grid. Clearly such insight is compatible
with the M/P approach (it is only an aide-memoire recording) and
use of them could (as here) avoid the move to candidate profiles.

I dislike having to use candidate profiles. I was introduced to them
in May last year when a Guardian (UK daily newspaper) article
described them (called "tiny writing") as the means of solution. It
was my primary method until I got bored with that and sought a
more supportive means of solution than going through the chore of
deriving candidates or else overloading short term memory. Thus
was the concept of mandatory pairs born!

However, I readily admit that I am prone to "give up" too easily
when resorting to candidate profiles. One of my ideas would be to
list the Hard/v.hard archives with whether or not I used just M/P
or had to transfer. This would give those tackling the archives an
additional measure - to match me by solving using M/Pairs alone
or to outplay me by solving without using candidate profiles. Of
course the true 'sticklers' would beat me every time!

Thank you for your interest. It is good that we are not just working
on the leading edge of ultra-hard puzzles but also working on the
solution of less hard puzzles with minimal resources.